Improve part of code using a loop

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Ancalagon8
Ancalagon8 2023-2-4
评论： Star Strider 2023-2-23
I have an annual table X, where I want to calculate the five maximum values for each month. My code works but I need to also apply it for other variables so I am wondering how to shrink it a bit using a loop. So far i splitted the initial annual table (X) into months like this:
X_January = table(Dates{1,1},X{1,1});
X_February = table(Dates{2,1},X{2,1});
X_March = table(Dates{3,1},X{3,1});
X_April = table(Dates{4,1},X{4,1});
X_May = table(Dates{5,1},X{5,1});
X_June = table(Dates{6,1},X{6,1});
X_July = table(Dates{7,1},X{7,1});
X_August = table(Dates{8,1},X{8,1});
X_September = table(Dates{9,1},X{9,1});
X_October = table(Dates{10,1},X{10,1});
X_November = table(Dates{11,1},X{11,1});
X_December = table(Dates{12,1},X{12,1});
%Max 5 values for January
[Col,idx] = maxk(X_January{:,1},5);
T= X_January(idx,:)
Any ideas how to make a loop?
  2 个评论
Dyuman Joshi
Dyuman Joshi 2023-2-4
编辑：Dyuman Joshi 2023-2-4
How do you want to store your final output?
All values in a numeric array (12x5 or 5x12 )? If otherwise, please specify.
Ancalagon8
Ancalagon8 2023-2-4
@Dyuman Joshi Yes, all values in a numeric array (12X5).

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Star Strider
Star Strider 2023-2-4
Ran in:
One approach —
date_time = (datetime('01-Jan-2023') : caldays(1) : datetime('31-Dec-2023')).';
Values = randn(numel(date_time), 5);
T1 = [table(date_time) array2table(Values)]
T1 = 365×6 table
date_time Values1 Values2 Values3 Values4 Values5 ___________ ________ ________ ________ _________ _________ 01-Jan-2023 -0.47138 0.78525 0.15011 0.83276 -0.59465 02-Jan-2023 -0.71248 -1.1215 0.2873 -0.07 0.48487 03-Jan-2023 1.4347 1.3957 -0.86495 0.38031 -1.3195 04-Jan-2023 0.90204 0.22882 0.52142 0.16887 -0.5348 05-Jan-2023 -0.52903 0.25618 1.0727 -1.0713 -1.1119 06-Jan-2023 0.41189 0.47979 -0.51813 -1.2685 1.2286 07-Jan-2023 -1.7744 -0.62537 -0.49037 -0.30792 -1.0039 08-Jan-2023 2.4883 -1.7341 0.48573 0.21777 -1.5177 09-Jan-2023 -0.97144 -0.19991 0.50003 -0.78522 -0.098178 10-Jan-2023 1.7618 0.17462 0.083165 -1.3276 0.80792 11-Jan-2023 -0.21396 -0.42347 0.69142 0.30005 0.16739 12-Jan-2023 1.2937 2.0955 -0.52163 -0.023047 1.8627 13-Jan-2023 0.025689 -1.0821 0.81364 1.0807 1.1739 14-Jan-2023 0.25328 -0.25033 -2.0148 0.25242 0.1674 15-Jan-2023 -0.87928 -0.26322 1.5713 0.16888 -0.88458 16-Jan-2023 -0.72874 0.11104 0.61025 2.2831 0.64283
for k = 1:12
Lv = month(T1.date_time) == k;
M = T1(Lv,:);
[~,idx] = maxk(M{:,2},5); % Index Returns 5 Highest Values Of First Variable
Month{k} = M(idx,:);
end
Month{1} % January
ans = 5×6 table
date_time Values1 Values2 Values3 Values4 Values5 ___________ _______ _______ ________ _________ ________ 08-Jan-2023 2.4883 -1.7341 0.48573 0.21777 -1.5177 10-Jan-2023 1.7618 0.17462 0.083165 -1.3276 0.80792 17-Jan-2023 1.4461 0.69264 0.47425 0.53047 -0.08912 03-Jan-2023 1.4347 1.3957 -0.86495 0.38031 -1.3195 12-Jan-2023 1.2937 2.0955 -0.52163 -0.023047 1.8627
Month{12} % December
ans = 5×6 table
date_time Values1 Values2 Values3 Values4 Values5 ___________ _______ ________ ________ ________ ________ 30-Dec-2023 1.3279 1.4798 -0.57554 -1.1291 0.14661 01-Dec-2023 1.0922 0.070399 0.1188 0.46017 -0.16276 18-Dec-2023 1.0347 -0.8301 1.3053 -0.11941 0.96864 21-Dec-2023 1.0141 -1.0614 2.0416 0.016393 -0.35875 24-Dec-2023 0.87875 -1.1881 -0.10361 1.5908 1.2401
Please do not use numbered variables! Simply store the results in an array.
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  14 个评论
显示 12更早的评论
Ancalagon8
Ancalagon8 2023-2-23
Thank you for your answer, but I meant an empty row (not collumn). Is it possible?
Star Strider
Star Strider 2023-2-23
Ran in:
As always, my pleasure!
The easiest way to find out is to do the experiment, adding:
Month{k} = [Month{k}; []];
That will not throw an error, however it does not affect the result, and neither does:
Month{k} = [Month{k}; cell(1,size(Month{k},2))];
while this works:
Month{k} = [Month{k}; table(NaT,NaN,NaN,NaN, 'VariableNames',Month{k}.Properties.VariableNames)];
however this (and analogues of it) do not:
Month{k} = [Month{k}; table({},[],[],[], 'VariableNames',Month{k}.Properties.VariableNames)];
So it appears not to be possible.
I will leave this up here so you can experiment with it —
date_time = (datetime('01-Jan-2023') : caldays(1) : datetime('31-Dec-2023')).';
Temperature = randn(numel(date_time), 1);
T1 = [table(date_time) array2table(Temperature)]
T1 = 365×2 table
date_time Temperature ___________ ___________ 01-Jan-2023 -1.462 02-Jan-2023 0.29838 03-Jan-2023 -1.2412 04-Jan-2023 0.040549 05-Jan-2023 0.25528 06-Jan-2023 -0.63519 07-Jan-2023 -1.6557 08-Jan-2023 -0.31906 09-Jan-2023 -0.73945 10-Jan-2023 0.34245 11-Jan-2023 0.31167 12-Jan-2023 1.4 13-Jan-2023 0.89639 14-Jan-2023 0.16224 15-Jan-2023 0.22773 16-Jan-2023 -0.16861
VN = T1.Properties.VariableNames;
[Y,M,D] = ymd(T1.date_time);
DN = day(T1.date_time,'dayofyear');
T1 = addvars(T1,M,Y,'AFter','date_time');
T1.Properties.VariableNames = {VN{1},'M','Y',VN{2}}; % Necessary, Since They Show Up As 'Var2' etc. Otherwise
T1
T1 = 365×4 table
date_time M Y Temperature ___________ _ ____ ___________ 01-Jan-2023 1 2023 -1.462 02-Jan-2023 1 2023 0.29838 03-Jan-2023 1 2023 -1.2412 04-Jan-2023 1 2023 0.040549 05-Jan-2023 1 2023 0.25528 06-Jan-2023 1 2023 -0.63519 07-Jan-2023 1 2023 -1.6557 08-Jan-2023 1 2023 -0.31906 09-Jan-2023 1 2023 -0.73945 10-Jan-2023 1 2023 0.34245 11-Jan-2023 1 2023 0.31167 12-Jan-2023 1 2023 1.4 13-Jan-2023 1 2023 0.89639 14-Jan-2023 1 2023 0.16224 15-Jan-2023 1 2023 0.22773 16-Jan-2023 1 2023 -0.16861
for k = 1:12
Lv = month(T1.date_time) == k;
M = T1(Lv,:);
[~,idx] = maxk(M{:,2},5); % Index Returns 5 Highest Values Of First Variable
Month{k} = M(idx,:);
% Q = cell(1,size(Month{k},2))
Month{k} = [Month{k}; table({},[],[],[], 'VariableNames',Month{k}.Properties.VariableNames)];
% Month{k} = [Month{k}; table(NaT,NaN,NaN,NaN, 'VariableNames',Month{k}.Properties.VariableNames)];
end
Month{1} % January
ans = 5×4 table
date_time M Y Temperature ___________ _ ____ ___________ 01-Jan-2023 1 2023 -1.462 02-Jan-2023 1 2023 0.29838 03-Jan-2023 1 2023 -1.2412 04-Jan-2023 1 2023 0.040549 05-Jan-2023 1 2023 0.25528
Month{12} % December
ans = 5×4 table
date_time M Y Temperature ___________ __ ____ ___________ 01-Dec-2023 12 2023 0.23843 02-Dec-2023 12 2023 1.5669 03-Dec-2023 12 2023 -0.48915 04-Dec-2023 12 2023 0.74177 05-Dec-2023 12 2023 -0.47897
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